Two photons of different frequencies collide to create electron and positron A photon of frequency f, and another of frequency f' (take f' as given) collide to create an electron-positron pair. The frequency f is such that when the collision is head on, there is exactly enough energy to produce the electron-positron pair at rest in the center-of-mass frame. What is f)
So in the rest frame, $E=h(f+f'), p=h\frac{f-f'}{c}$. We know that in the center-of-mass frame, $E'=2m_ec^2$. So my idea was to find the velocity of the center of mass frame, and then $E=\gamma E'$, but this seems overly complicated, is there an easier way?
 A: Using the conservation of energy, we can write $$h(f+f')=2m_{e}c^{2} +T_{e+}+T_{e-}$$ where $T_{e+}$ is the kinetic energy in the positron and $T_{e-}$ is the kinetic energy of the electron. These values are zero for your case, if I'm reading it correctly. If the collision is head on, this makes this much simpler. We can identify $2m_{e}c^{2}$ as the minimum energy needed for Pair Production. The question seems to tell us that you then have $$h(f+f')=1.022 ~ \text  MeV$$If the frequency is such that when hit head on there is enough energy for pair production then we have a case where the photon of frequency $f$ has the minimum energy needed for this process and hence has the maximum wavelength. $$\frac{hc}{\lambda_{max}}=2m_{e}c^{2}$$ So now we can say $$\lambda_{max}=\frac{h}{2m_{e}c}=1.21 \cdot 10^{-12}~\text m$$
Please let me know if I'm unclear. Hope this helps! Cheers
A: In the center of mass frame, the frequency of each photon is $f''=m_e c^2/h$. Due to the relativistic Doppler effect, $f/f''=f''/f'$, if I am not mistaken. 
